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3 years ago

Solve the system of equations 2x+2y=6z=8 5x+6y+5z=3 5x+6y+3z=5

Mathematics

1 answer:

Alborosie3 years ago

4 0

Answer:

$x=34,y=-27andz=-1$

Step-by-step explanation:

Given:

$2x+2y+6z=8$ equation:1

$5x+6y+5z=3$ equation:2

$5x+6y+3z=5$ equation:3

We have to find the value of x,y and z

Eliminating two variables to find the value of third variable

Subtracting equation:2 from equation:3

$5x+6y+3z-(5x+6y+5z)=5-3$

$5x+6y+3z-5x-6y-5z=2\\-2z=2\\z=-1$

Solving the equation:1 for 'x'

$2x+2y+6z=8$

$2x=8-2y-6z\\2x=8-2y-6(-1)\\2x=8-2y+6\\2x=14-2y\\x=7-y$

Putting x and z in equation 2

$5(7-y)+6y+5(-1)=3\\35-5y+6y-5=3\\y+30=3\\y=3-30\\y=-27$

Putting the value of 'y' and 'z' in equation:2 to find the value of 'x'

$5x+6(-27)+5(-1)=3\\5x-162-5=3\\5x-167=3\\5x=3+167\\5x=170\\x=\frac{170}{5}\\ x=34$

The values of x,y and z are

$x=34,y=-27andz=-1$

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Read 2 more answers

Using the 0,1,6,8, and 9, write down all the possible three-digit numbers that have rotational symmetry. the digits can be used

Hunter-Best [27]

Assuming 1 is written rotationally symmetrical, using 0,1,6,8,9 to make numbers that have a degree of symmetry of 2 (i.e. 180 degrees), we can say

one digit numbers: 0,1,8

two digit numbers: 11, 69, 96

(18 becomes 81, so not admissible, similarly for 19, 91,16,61, etc)

(00 is not considered generally as a two-digit number)

three digit numbers:

(000, 010, 080 are not considered generally as a three-digit number)

101, 609, 808, 906,

replace the middle digit successively by 1 and 8 gives

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181, 689, 888, 986

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3 years ago

Express 3080 as a product of prime factors

n200080 [17]

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Which is the same as = 23 x 5 x 7 x 11

Prime Factors Tree of 3080

3080
/ \
2 1540
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3 years ago

Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv

ch4aika [34]

Answer:

Part 1) The function of the First graph is $f(x)=(x-3)(x+1)$

Part 2) The function of the Second graph is $f(x)=-2(x-1)(x+3)$

Part 3) The function of the Third graph is $f(x)=0.5(x-6)(x+2)$

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

$f(x)=a(x-c)(x-d)$

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-3)(x+1)$

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

$-4=a(1-3)(1+1)$

$-4=a(-2)(2)$

$a=1$

therefore

The function is equal to

$f(x)=(x-3)(x+1)$

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

$f(x)=a(x-1)(x+3)$

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

$8=a(-1-1)(-1+3)$

$8=a(-2)(2)$

$a=-2$

therefore

The function is equal to

$f(x)=-2(x-1)(x+3)$

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-6)(x+2)$

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

$-8=a(2-6)(2+2)$

$-8=a(-4)(4)$

$a=0.5$

therefore

The function is equal to

$f(x)=0.5(x-6)(x+2)$

3 0

3 years ago